# Angular Momentum

Angular Momentum

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• Angular momentum is the tendency of a spinning object to continue to spin.

• Angular momentum measures the amount of "spinning" of a system. So the conservation of angular momentum says that, in the absence of a net torque, the amount of "spinning" in a system is the same at all times.

I'm using the word "spinning" really loosely here -- really, angular momentum is composed of the revolution of objects around an axis, and the rotation of other objects about the axis. Objects that revolve contribute to orbital angular momentum, and objects that rotate contribute to spin angular momentum. Even though it's composed of these two parts, I lack a better word than "spinning" to describe what angular momentum measures, so I'll just stick with that.

So if you go out and measure the amount of "spinning" of a system in which angular momentum is conserved, you'll find that no matter when you measure it, it will always be the same.

I also mentioned torques briefly above, so I might as well talk about them briefly. Torque is to angular momentum as force is to linear momentum ($\vec{p} = m \vec{v}$). Torques are the rates of change of angular momenta, so in systems where there is a net torque, angular momentum is not conserved.

• Angular momentum depends on the angular velocity and moment of inertia. We have the equation

L = Iω

where $L$ is angular momentum, $I$ is the moment of inertia, and ω is angular velocity.

For example: consider a thin rod (like a pencil) rotating around one if it's ends. You can try this by sticking your pencil to a needle by it's eraser.

The moment of inertia for a thin rod rotating about one its end is given by: $I = \frac{m {l}^{2}}{3}$ (finding why this is the moment of inertia is another exercise -- but an important one). Here the length of the rod is $l$ and it's total mass is $m$. The moment of inertia is different for different objects/shapes.

If you give it an angular velocity $\omega$, the rod will then have angular momentum:

$L = \frac{m {l}^{2} \omega}{3}$